Suppose that the length of long distance phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to $10 minutes. a. The lambda of this distribution is b. The probability that the length of a phone call is longer than 12 is P(x 2 12) = c. The probability that the length of a phone call is shorter than 5 is P(x s 5) = d. The probability that the length of a phone call is between 8 and 14 is P(8 s x s 14) =| e. The 86th percentile is a phone call that lasts minutes.

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Suppose that the length of long distance phone calls, measured in minutes, is known to have an
exponential distribution with the average length of a call equal to $10 minutes.
a. The lambda of this distribution is
b. The probability that the length of a phone call is longer than 12 is P(x 2 12) =
c. The probability that the length of a phone call is shorter than 5 is P(x s 5) =
d. The probability that the length of a phone call is between 8 and 14 is P(8 s x s 14) =
e. The 86th percentile is a phone call that lasts
minutes.
Transcribed Image Text:Suppose that the length of long distance phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to $10 minutes. a. The lambda of this distribution is b. The probability that the length of a phone call is longer than 12 is P(x 2 12) = c. The probability that the length of a phone call is shorter than 5 is P(x s 5) = d. The probability that the length of a phone call is between 8 and 14 is P(8 s x s 14) = e. The 86th percentile is a phone call that lasts minutes.
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a) Given that the average length of a call is 10 minutes.This means 1λ=10 or λ=110=0.1

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